# How do I find the bounds for an integral between a plane and a sphere? So this is what I got so far:

$$\int _0^{2\pi }\:\int _0^{\sqrt{5}}\:\int _3^{\sqrt{16-r^2}}\:\left(r\right)dzdrd\theta$$

And the answer is shown above in the screenshot. I'm not exactly sure which bounds I'm messing up.

I know it's a sphere, so it's gonna be $$0$$ to $$2\pi$$ for theta. The middle one I plug in $$z=3$$ and subtract and replace the squared $$x$$ and $$y$$ with $$r$$ squared. And then the furthest right one I do the same thing pretty much there.

What's my problem? Thank you.

Perhaps it would be easier to start by finding the volume of a small slice of the sphere created by leaving only the portion of the sphere that lies in the range $$z=a\dots b,\,b>a$$.
This section would essentially be a flat pancake with its symmetry axis along z-axis. What is the radius of that pancake? Call it $$\rho$$. From basic Pythagoras theorem $$\rho=\sqrt{r^2-z^2}$$, where $$r$$ is the radius of the sphere. Clearly this only makes sense for $$z^2\le r^2$$.
So now you can find the volume of this pancake (note how positioning of $$\int$$ and $$d\dots$$ can be used to clarify the bounds in the integrals):
$$V_{pancake}\left(a,b,r\right)=\int_a^b dz\int_0^\sqrt{r^2-z^2} \rho d\rho\int_0^{2\pi} d\phi=2\pi\int_a^b dz \frac{\left(\sqrt{r^2-z^2}\right)^2}{2}=\dots$$
• @Dr.SuessOfficial, it's at the beginning of my reply. Consider a sphere now remove the portion of the sphere below $z<a$ and above $z>b$. Provided $a$ and $b$ are close, you will be left with a flat pancake. Then you find the volume of that pancake. The final expression I gave above is valid even if the pancake is not thin, i.e. for any $-r\le a\le b\le r$ – Cryo Oct 30 at 21:55